Welcome to Lecture 102 of our comprehensive JEE Physics Master Series! Continuing our study in Unit 9: Kinetic Theory of Gases, we explore a core thermodynamic mechanism with profound engineering and exam relevance: Work Done on Compressing a Gas. When an external agent compresses a gas, mechanical energy is transferred into the system, altering its pressure, volume, temperature, and internal energy. We contrast quasi-static reversible compression across isobaric, isothermal, adiabatic, and polytropic paths, explain why isothermal compression requires the absolute minimum work, examine sudden single-stage and multi-stage irreversible compressions, derive the optimal pressure ratio for multi-stage compression with intercooling, and solve standard JEE Advanced problem archetypes.
Detailed Concept Card: Work Done on Compressing a Gas
1. Sign Conventions & General Definition of Compressive Work
In classical physics and thermodynamics, work done by the gas on its surroundings during a volume change from $V_1$ to $V_2$ is defined by:
$$W_{\text{by gas}} = \int_{V_1}^{V_2} P_{\text{ext}} \, dV$$
In a compression process, the final volume is strictly smaller than the initial volume ($V_2 < V_1 \implies dV < 0$). Consequently:
- Work done BY the gas is negative: $W_{\text{by gas}} < 0$.
- Work done ON the gas by the external agent is positive: $$\mathbf{W_{\text{on gas}} = -W_{\text{by gas}} = -\int_{V_1}^{V_2} P_{\text{ext}} \, dV = \int_{V_2}^{V_1} P_{\text{ext}} \, dV > 0}$$
- For a quasi-static reversible compression, the external pressure equals the internal gas pressure at every instant ($P_{\text{ext}} = P_{\text{gas}} = P$): $$\mathbf{W_{\text{on gas}} = -\int_{V_1}^{V_2} P \, dV = \int_{V_2}^{V_1} P \, dV}$$ Graphically, $W_{\text{on gas}}$ represents the area under the process curve projected onto the volume axis taken from $V_2$ to $V_1$ on a $P$-$V$ indicator diagram.
2. Reversible Compression Paths (Fixed Volume Ratio $r = V_1 / V_2 > 1$)
Consider $n\text{ moles}$ of an ideal gas initially at state $(P_1, V_1, T_1)$ compressed reversibly to final volume $V_2$ ($V_2 < V_1$) along different thermodynamic pathways:
| Compression Path | Governing Equation | Final State Coordinates ($P_2, T_2$) | Work Done ON the Gas ($W_{\text{on}}$) |
|---|---|---|---|
| Isobaric Compression | $P = P_1 = \text{const}$ | $$P_2 = P_1, \quad T_2 = T_1 \left(\frac{V_2}{V_1}\right) < T_1$$ | $$W_{\text{on}} = P_1 (V_1 – V_2) = n R (T_1 – T_2)$$ |
| Isothermal Compression | $P V = n R T_1 = \text{const}$ | $$P_2 = P_1 \left(\frac{V_1}{V_2}\right) > P_1, \quad T_2 = T_1$$ | $$W_{\text{on}} = n R T_1 \ln\left(\frac{V_1}{V_2}\right) = P_1 V_1 \ln\left(\frac{P_2}{P_1}\right)$$ |
| Polytropic Compression ($1 < m < \gamma$) | $P V^m = \text{const}$ | $$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^m, \quad T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{m-1}$$ | $$W_{\text{on}} = \frac{P_2 V_2 – P_1 V_1}{m – 1} = \frac{n R (T_2 – T_1)}{m – 1}$$ |
| Reversible Adiabatic Compression | $P V^\gamma = \text{const}$ ($Q = 0$) | $$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma, \quad T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} > T_1$$ | $$W_{\text{on}} = \frac{P_2 V_2 – P_1 V_1}{\gamma – 1} = \frac{n R (T_2 – T_1)}{\gamma – 1} = \Delta U$$ |
Comparison of Work Required for a Fixed Volume Reduction ($V_1 \to V_2$)
On a $P$-$V$ indicator diagram, moving from initial volume $V_1$ leftward to smaller volume $V_2$:
- The slope magnitude of an adiabatic curve is $\gamma$ times steeper than that of an isothermal curve: $\left|\frac{dP}{dV}\right|_{\text{adia}} = \gamma \frac{P}{V} > \left|\frac{dP}{dV}\right|_{\text{iso}} = \frac{P}{V}$.
- Therefore, during compression, the adiabatic curve climbs above the polytropic curve, which climbs above the isothermal curve, which stays above the isobaric line.
- Comparing the areas under the curves from $V_1$ to $V_2$: $$\mathbf{W_{\text{on, adia}} > W_{\text{on, poly}} > W_{\text{on, iso}} > W_{\text{on, isobaric}}}$$
- Physical Insight: In adiabatic compression, no heat is removed ($Q = 0$), so all work done on the gas converts directly into internal thermal energy, raising the temperature ($T_2 > T_1$). The higher temperature produces higher pressure, requiring greater external force and work at every subsequent millimeter of compression. In isothermal compression, continuous heat removal ($Q = W_{\text{by}} < 0$) keeps the gas cool, minimizing pressure buildup and minimizing the total compressive work required.
3. Compression to the Same Final Pressure ($P_1 \to P_2$)
When compressing an ideal gas from initial pressure $P_1$ to a specified delivery pressure $P_2$ ($P_2 > P_1$ with pressure ratio $r_p = P_2 / P_1$):
- Final Volume in Isothermal Compression: $V_{2,\text{iso}} = \frac{V_1}{r_p}$.
- Final Volume in Adiabatic Compression: $V_{2,\text{adia}} = \frac{V_1}{r_p^{1/\gamma}} > V_{2,\text{iso}}$ (since $\gamma > 1 \implies r_p^{1/\gamma} < r_p$).
- Work Done Comparison: $$W_{\text{on, iso}} = P_1 V_1 \ln\left(\frac{P_2}{P_1}\right)$$ $$W_{\text{on, adia}} = \frac{P_2 V_{2,\text{adia}} – P_1 V_1}{\gamma – 1} = \frac{P_1 V_1}{\gamma – 1}\left[\left(\frac{P_2}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} – 1\right]$$ For any pressure ratio $r_p > 1$, $\mathbf{W_{\text{on, adia}} > W_{\text{on, iso}}}$. Isothermal compression remains the thermodynamic benchmark for maximum energy efficiency.
4. Irreversible Sudden Compression: Single-Stage vs. Multi-Stage
In practical reciprocating machinery, compression is often rapid and non-quasi-static:
- Single-Stage Sudden Isothermal Compression:
A gas at $(P_1, V_1)$ is suddenly subjected to constant external pressure $P_{\text{ext}} = P_2. The piston compresses the gas until equilibrium is reached at $V_2 = \frac{P_1 V_1}{P_2}$: $$\mathbf{W_{\text{on, single}} = P_2 (V_1 – V_2) = P_2 \left(V_1 – \frac{P_1 V_1}{P_2}\right) = P_1 V_1 \left(\frac{P_2}{P_1} – 1\right)}$$ Since $r_p – 1 > \ln r_p$ for all $r_p > 1$, irreversible single-stage compression requires significantly more work than reversible compression ($W_{\text{on, single}} > W_{\text{on, rev}}$). - Two-Stage Sudden Isothermal Compression:
If the compression is executed in two steps via an intermediate pressure $P_i$ ($P_1 < P_i < P_2$): $$W_{\text{on, 2-stage}} = P_i (V_1 - V_i) + P_2 (V_i - V_2) = P_1 V_1 \left(\frac{P_i}{P_1} - 1\right) + P_i V_i \left(\frac{P_2}{P_i} - 1\right) = P_1 V_1 \left(\frac{P_i}{P_1} + \frac{P_2}{P_i} - 2\right)$$ To minimize $W_{\text{on, 2-stage}}$, differentiate with respect to $P_i$: $$\frac{d}{dP_i}\left(\frac{P_i}{P_1} + \frac{P_2}{P_i}\right) = \frac{1}{P_1} - \frac{P_2}{P_i^2} = 0 \implies \mathbf{P_i = \sqrt{P_1 P_2}}$$ Theorem: The optimal intermediate pressure is the geometric mean of the initial and final pressures. Under this optimal condition, the pressure ratios in both stages are identical ($\frac{P_i}{P_1} = \frac{P_2}{P_i} = \sqrt{\frac{P_2}{P_1}}$), and the work done in each stage is equal!
5. Multi-Stage Compression with Perfect Intercooling
To avoid the excessive temperature rise and heavy work penalties of high-pressure adiabatic compressors, engineering designs employ multi-stage compression with intercoolers:
- Stage 1: Gas is compressed adiabatically from $(P_1, T_1)$ to an intermediate pressure $P_i$, heating to temperature $T_{\text{exit}, 1} = T_1 \left(\frac{P_i}{P_1}\right)^{\frac{\gamma – 1}{\gamma}}$.
- Intercooler: The gas passes through an isobaric heat exchanger at constant pressure $P_i$, cooling back down to the initial temperature $T_1$ (perfect intercooling).
- Stage 2: Gas enters the second cylinder at $(P_i, T_1)$ and is compressed adiabatically to final pressure $P_2$.
The total work required for a two-stage compressor with perfect intercooling is:
$$W_{\text{total}} = \frac{n R T_1 \gamma}{\gamma – 1} \left[\left(\frac{P_i}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} + \left(\frac{P_2}{P_i}\right)^{\frac{\gamma – 1}{\gamma}} – 2\right]$$
Minimizing $W_{\text{total}}$ with respect to $P_i$ yields the foundational result:
$$\mathbf{P_i = \sqrt{P_1 P_2} \iff \frac{P_i}{P_1} = \frac{P_2}{P_i} = \left(\frac{P_2}{P_1}\right)^{1/2}}$$
For an $N$-stage compressor with $(N – 1)$ perfect intercoolers, minimum work requires equal pressure ratios across all stages:
$$\mathbf{r_{p, \text{stage}} = \left(\frac{P_{\text{final}}}{P_{\text{initial}}}\right)^{1/N}, \quad W_{\text{stage 1}} = W_{\text{stage 2}} = \dots = W_{\text{stage N}}}$$
Four Graded Solved Examples
Example 1 (JEE Main Level: Isothermal vs. Adiabatic Compression to One-Eighth Volume)
Problem Statement: One mole of an ideal monoatomic gas ($\gamma = 5/3, C_v = 1.5R$) initially at pressure $P_0$, volume $V_0$, and temperature $T_0 = 300\text{ K}$ is compressed to one-eighth of its initial volume ($V_f = V_0 / 8$). Calculate the final pressure, final temperature, and work done ON the gas in terms of $R T_0$ for:
(a) Reversible isothermal compression.
(b) Reversible adiabatic compression.
Given $\ln 2 = 0.693$, find the ratio of adiabatic work to isothermal work.
Solution:
Part (a): Reversible Isothermal Compression ($T = T_0 = \text{const}$).
Final pressure by Boyle’s Law:
$$P_{\text{iso}} = P_0 \left(\frac{V_0}{V_0 / 8}\right) = \mathbf{8 P_0}$$
Final temperature:
$$T_{\text{iso}} = \mathbf{T_0 = 300\text{ K}}$$
Work done ON the gas:
$$W_{\text{on, iso}} = n R T_0 \ln\left(\frac{V_0}{V_f}\right) = (1) R T_0 \ln(8) = 3 R T_0 \ln 2 = 3(0.693) R T_0 = \mathbf{2.079 \, R T_0}$$
Part (b): Reversible Adiabatic Compression ($P V^{5/3} = \text{const}$).
Final pressure by Poisson’s Law:
$$P_{\text{adia}} = P_0 \left(\frac{V_0}{V_f}\right)^\gamma = P_0 (8)^{5/3} = P_0 (2^3)^{5/3} = P_0 (2^5) = \mathbf{32 P_0}$$
Final temperature from $T V^{\gamma – 1} = \text{const}$:
$$T_{\text{adia}} = T_0 \left(\frac{V_0}{V_f}\right)^{\gamma – 1} = T_0 (8)^{5/3 – 1} = T_0 (8)^{2/3} = T_0 (2^3)^{2/3} = 4 T_0 = \mathbf{1200\text{ K}}$$
Work done ON the gas equals the gain in internal energy ($\Delta U$):
$$W_{\text{on, adia}} = \frac{P_{\text{adia}} V_f – P_0 V_0}{\gamma – 1} = \frac{(32 P_0)(V_0 / 8) – P_0 V_0}{5/3 – 1} = \frac{4 P_0 V_0 – P_0 V_0}{2/3} = \frac{3 P_0 V_0}{2/3} = \mathbf{4.50 \, P_0 V_0 = 4.50 \, R T_0}$$
Part (c): Comparison of Work Required.
$$\frac{W_{\text{on, adia}}}{W_{\text{on, iso}}} = \frac{4.50 \, R T_0}{2.079 \, R T_0} \approx \mathbf{2.164}$$
Adiabatic compression requires more than double the work of isothermal compression because internal heating dramatically elevates the resisting pressure ($32 P_0$ vs. $8 P_0$).
Example 2 (JEE Main/Advanced: Single-Stage vs. Two-Stage Sudden Irreversible Compression)
Problem Statement: One mole of an ideal gas at temperature $T_0 = 300\text{ K}$ and initial pressure $P_1 = 1.0\text{ atm}$ is compressed isothermally to a final pressure $P_2 = 4.0\text{ atm}$. Given $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $\ln 2 = 0.693$, compare the work done on the gas in three scenarios:
1. Scenario A: Sudden single-stage compression against constant external pressure $P_{\text{ext}} = P_2 = 4.0\text{ atm}$.
2. Scenario B: Sudden two-stage compression using the optimal intermediate pressure $P_i = \sqrt{P_1 P_2} = 2.0\text{ atm}$.
3. Scenario C: Ideal reversible isothermal compression.
Find the percentage work saved by adopting two-stage compression over single-stage compression.
Solution:
Note that $R T_0 = (8.314)(300) = 2494.2\text{ J}$.
Scenario A: Single-Stage Sudden Compression ($P_{\text{ext}} = 4.0\text{ atm}$).
$$W_{\text{on, single}} = P_2 (V_1 – V_2) = P_2 \left(\frac{R T_0}{P_1} – \frac{R T_0}{P_2}\right) = R T_0 \left(\frac{P_2}{P_1} – 1\right)$$
$$W_{\text{on, single}} = R T_0 (4.0 – 1) = 3 R T_0 = 3 \times 2494.2\text{ J} = \mathbf{7482.6\text{ J}}$$
Scenario B: Two-Stage Sudden Compression ($P_i = 2.0\text{ atm}, P_2 = 4.0\text{ atm}$).
In Stage 1 (compressed from $P_1$ to $P_i$):
$$W_1 = P_i (V_1 – V_i) = R T_0 \left(\frac{P_i}{P_1} – 1\right) = R T_0 (2.0 – 1) = 1 R T_0$$
In Stage 2 (compressed from $P_i$ to $P_2$):
$$W_2 = P_2 (V_i – V_2) = R T_0 \left(\frac{P_2}{P_i} – 1\right) = R T_0 (2.0 – 1) = 1 R T_0$$
Total two-stage work:
$$W_{\text{on, 2-stage}} = W_1 + W_2 = 1 R T_0 + 1 R T_0 = 2 R T_0 = 2 \times 2494.2\text{ J} = \mathbf{4988.4\text{ J}}$$
Scenario C: Reversible Isothermal Compression.
$$W_{\text{on, rev}} = R T_0 \ln\left(\frac{P_2}{P_1}\right) = R T_0 \ln(4) = 2 R T_0 \ln 2 = 2(0.693)(2494.2\text{ J}) = \mathbf{3457.0\text{ J}}$$
Percentage Work Saved by Two-Stage Operation:
$$\text{Savings} = \frac{W_{\text{on, single}} – W_{\text{on, 2-stage}}}{W_{\text{on, single}}} \times 100\% = \frac{7482.6 – 4988.4}{7482.6} \times 100\% = \frac{2494.2}{7482.6} \times 100\% = \mathbf{33.33\%}$$
Dividing the compression into just two equal pressure-ratio stages saves one-third of the expended work!
Example 3 (JEE Advanced Level: Two-Stage Compressor with Perfect Intercooling)
Problem Statement: Air ($\gamma = 1.4 = 7/5, C_p = 3.5R$) enters an industrial compressor at $P_1 = 100\text{ kPa}$ and $T_1 = 300\text{ K}$, and is compressed to delivery pressure $P_2 = 900\text{ kPa}$. The compression is carried out in two reversible adiabatic stages with an intercooler between them that cools the air at constant pressure back to $300\text{ K}$. Given $3^{2/7} \approx 1.3687$ and $9^{2/7} \approx 1.8734$:
1. Determine the optimal intermediate pressure $P_i$ (in $\text{kPa}$) that minimizes the total compression work.
2. Calculate the maximum temperature reached by the air in each stage.
3. Compare the total work with that required by a single-stage adiabatic compressor operating between $100\text{ kPa}$ and $900\text{ kPa}$.
Solution:
Step 1: Optimal Intermediate Pressure.
For minimum total work with perfect intercooling:
$$P_i = \sqrt{P_1 P_2} = \sqrt{(100\text{ kPa})(900\text{ kPa})} = \sqrt{90000} = \mathbf{300\text{ kPa}}$$
The pressure ratio in each stage is equal:
$$r_p = \frac{P_i}{P_1} = \frac{300}{100} = 3.0, \quad \frac{P_2}{P_i} = \frac{900}{300} = 3.0$$
Step 2: Exit Temperature in Each Stage.
In Stage 1 ($100\text{ kPa} \to 300\text{ kPa}$):
$$T_{\text{exit}, 1} = T_1 \left(\frac{P_i}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} = 300 \times 3^{0.4 / 1.4} = 300 \times 3^{2/7} = 300 \times 1.3687 = \mathbf{410.6\text{ K}}$$
The gas is then cooled isobarically in the intercooler from $410.6\text{ K}$ back to $T_1 = 300\text{ K}$.
In Stage 2 ($300\text{ kPa} \to 900\text{ kPa}$ from initial temperature $300\text{ K}$):
$$T_{\text{exit}, 2} = T_1 \left(\frac{P_2}{P_i}\right)^{\frac{\gamma – 1}{\gamma}} = 300 \times 3^{2/7} = \mathbf{410.6\text{ K}}$$
The maximum cycle temperature is safely capped at $410.6\text{ K}$ ($137.4^\circ\text{C}$).
Step 3: Work Comparison with Single-Stage Compression.
For two-stage compression with intercooling, work done per mole is:
$$W_{\text{2-stage}} = 2 \times n C_p (T_{\text{exit}} – T_1) = 2 \times (3.5 R)(410.6 – 300) = 7 R (110.6) = \mathbf{774.2 \, R}$$
For a single-stage adiabatic compressor without intercooling ($100\text{ kPa} \to 900\text{ kPa}$):
$$T_{\text{single}} = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} = 300 \times 9^{2/7} = 300 \times 1.8734 = \mathbf{562.0\text{ K}}$$
Single-stage work per mole:
$$W_{\text{single}} = n C_p (T_{\text{single}} – T_1) = 3.5 R (562.0 – 300) = 3.5 R (262.0) = \mathbf{917.0 \, R}$$
Work saved by two-stage intercooling:
$$\Delta W = W_{\text{single}} – W_{\text{2-stage}} = 917.0 R – 774.2 R = 142.8 \, R = 142.8 \times 8.314\text{ J/mol} \approx \mathbf{1187\text{ J/mol}}$$
Intercooling saves $15.6\%$ of the energy while preventing extreme thermal degradation ($562\text{ K} = 289^\circ\text{C}$).
Example 4 (JEE Advanced Level: Compressing Gas with an External Linear Spring)
Problem Statement: A vertical cylinder of cross-sectional area $A$ contains $n = 1\text{ mole}$ of an ideal gas at pressure $P_0$, volume $V_0$, and temperature $T_0$, sealed by a massless, frictionless piston. A vertical linear spring of spring constant $k$ is mounted above the piston, initially relaxed and just touching the top face of the piston. An external agent pushes the piston downward slowly, compressing the gas until its volume is halved ($V_f = V_0 / 2$). During this compression, the gas temperature is held constant at $T_0$ by contact with a thermal reservoir.
1. Write the expression for the external force $F_{\text{ext}}(x)$ exerted by the agent as a function of downward displacement $x$.
2. Calculate the total work $W_{\text{agent}}$ done by the external agent in compressing the gas and spring.
Solution:
Step 1: Free Body Diagram & External Force.
When the piston moves downward by displacement $x$, the gas volume decreases to:
$$V(x) = V_0 – A x$$
Since the gas is compressed isothermally at $T_0$:
$$P_{\text{gas}}(x) = \frac{P_0 V_0}{V(x)} = \frac{P_0 V_0}{V_0 – A x}$$
Forces acting on the massless piston in quasi-static equilibrium:
– Downward: External applied force $F_{\text{ext}}$ + atmospheric force $P_{\text{atm}} A$ (assume $P_{\text{atm}} = P_0$).
– Upward: Gas pressure force $P_{\text{gas}} A$ + spring restoring force $k x$.
Balancing vertical forces:
$$F_{\text{ext}} + P_0 A = P_{\text{gas}} A + k x \implies \mathbf{F_{\text{ext}}(x) = \left(\frac{P_0 V_0}{V_0 – A x} – P_0\right) A + k x}$$
Step 2: Calculate Work Done by External Agent.
The maximum downward displacement $x_f$ corresponding to $V_f = V_0 / 2$ is:
$$A x_f = V_0 – \frac{V_0}{2} = \frac{V_0}{2} \implies x_f = \frac{V_0}{2A}$$
The work performed by the agent is the integral of $F_{\text{ext}}$ from $x = 0$ to $x_f$:
$$W_{\text{agent}} = \int_0^{x_f} F_{\text{ext}}(x) \, dx = \int_0^{x_f} \left[\frac{P_0 V_0 A}{V_0 – A x} – P_0 A\right] dx + \int_0^{x_f} k x \, dx$$
Evaluating each integral:
1. Gas compression term:
$$\int_0^{x_f} \frac{P_0 V_0 A}{V_0 – A x} dx = -P_0 V_0 \left[\ln(V_0 – A x)\right]_0^{x_f} = -P_0 V_0 \ln\left(\frac{V_0 / 2}{V_0}\right) = P_0 V_0 \ln 2$$
2. Constant pressure term:
$$-\int_0^{x_f} P_0 A \, dx = -P_0 A x_f = -P_0 A \left(\frac{V_0}{2A}\right) = -\frac{1}{2} P_0 V_0$$
3. Spring energy term:
$$\int_0^{x_f} k x \, dx = \frac{1}{2} k x_f^2 = \frac{1}{2} k \left(\frac{V_0}{2A}\right)^2 = \frac{k V_0^2}{8 A^2}$$
Combining terms:
$$\mathbf{W_{\text{agent}} = P_0 V_0 \left(\ln 2 – \frac{1}{2}\right) + \frac{k V_0^2}{8 A^2}}$$
Takeaway: The agent supplies work to compress the isothermal gas against atmospheric backpressure ($P_0 V_0(\ln 2 – 0.5)$) plus the elastic potential energy stored in the spring ($\frac{k V_0^2}{8A^2}$).
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. An ideal gas is compressed to half its initial volume from the same initial equilibrium state $(P_0, V_0)$ along three different reversible paths: Path 1 is isobaric, Path 2 is isothermal, and Path 3 is adiabatic. If $W_1, W_2$, and $W_3$ denote the work done ON the gas in Paths 1, 2, and 3 respectively, then:
(A) $W_3 > W_2 > W_1$
(B) $W_1 > W_2 > W_3$
(C) $W_2 > W_3 > W_1$
(D) $W_3 > W_1 > W_2$
Q2. One mole of an ideal gas at temperature $T_0$ is compressed reversibly and isothermally until its volume decreases by a factor of $e$ ($V_f = V_0 / e$). The work done ON the gas during this process is:
(A) $R T_0$
(B) $-R T_0$
(C) $e R T_0$
(D) $\frac{R T_0}{e}$
Q3. An ideal rigid diatomic gas ($\gamma = 1.4$) initially at temperature $T_0 = 300\text{ K}$ is compressed reversibly and adiabatically until its volume is reduced to $\frac{1}{32}$ of its initial volume. The final temperature of the gas is:
(A) $600\text{ K}$
(B) $900\text{ K}$
(C) $1200\text{ K}$
(D) $2400\text{ K}$
Q4. In a two-stage air compressor with perfect intercooling between stages, air is drawn in at pressure $P_1$ and delivered at pressure $P_2$. The optimal intermediate pressure $P_i$ that minimizes the total compression work is:
(A) $\frac{P_1 + P_2}{2}$
(B) $\sqrt{P_1 P_2}$
(C) $\frac{2 P_1 P_2}{P_1 + P_2}$
(D) $\sqrt{P_1^2 + P_2^2}$
Q5. An ideal gas at initial pressure $P_1$ and volume $V_1$ is compressed suddenly to pressure $P_2 = 3 P_1$ by applying a constant external pressure $P_{\text{ext}} = 3 P_1$ in a single stage. If the temperature of the gas remains constant at $T_0$ after the compression, the work done ON the gas is:
(A) $P_1 V_1 \ln 3$
(B) $2 P_1 V_1$
(C) $3 P_1 V_1$
(D) $1.5 P_1 V_1$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Starting from the same initial state $(P_0, V_0, T_0)$, an ideal gas is compressed to final pressure $P_f = 2 P_0$ along an isothermal path (Process A) and along a reversible adiabatic path (Process B). Which of the following statements is/are correct?
(A) The final volume in Process A is smaller than in Process B ($V_{f, \text{A}} < V_{f, \text{B}}$).
(B) The magnitude of work done on the gas in Process B is greater than in Process A ($W_{\text{on, B}} > W_{\text{on, A}}$).
(C) The final temperature in Process B is higher than in Process A ($T_{f, \text{B}} > T_{f, \text{A}}$).
(D) On a $P$-$V$ diagram, the adiabatic curve lies entirely above the isothermal curve during the compression.
Q7. Which of the following statements regarding multi-stage gas compression with intercooling is/are correct?
(A) Intercooling reduces the total mechanical work required by keeping the average operating temperature lower.
(B) For a two-stage compressor with perfect intercooling, minimum work is achieved when the pressure ratio across both stages is identical.
(C) When optimized for minimum work, the work performed in the low-pressure stage equals the work performed in the high-pressure stage.
(D) Multi-stage compression with an infinite number of intercooled stages approaches reversible isothermal compression.
Q8. An ideal gas undergoes isothermal compression from $(P_1, V_1)$ to $(P_2, V_2)$ via two different methods: Method I is reversible quasi-static compression; Method II is a sudden single-stage irreversible compression against $P_{\text{ext}} = P_2$. Which of the following statements is/are correct?
(A) The change in internal energy of the gas is zero in both methods ($\Delta U_{\text{I}} = \Delta U_{\text{II}} = 0$).
(B) The heat rejected to the reservoir in Method II is greater than in Method I ($|Q_{\text{II}}| > |Q_{\text{I}}|$).
(C) The entropy change of the gas is identical in both methods ($\Delta S_{\text{gas, I}} = \Delta S_{\text{gas, II}} < 0$).
(D) The total entropy change of the universe in Method I is zero ($\Delta S_{\text{univ, I}} = 0$), whereas in Method II it is strictly positive ($\Delta S_{\text{univ, II}} > 0$).
Questions 9 to 10: Numerical / Integer Answer Type
Q9. One mole of an ideal monoatomic gas ($\gamma = 5/3$) initially at an energy state where $n R T_0 = 2000\text{ J}$ is compressed reversibly and adiabatically from volume $V_0$ to volume $V_0 / 8$. Determine the total work done ON the gas (in $\text{J}$).
Q10. An industrial compressor takes in air at $P_1 = 1.0\text{ bar}$ and delivers it at $P_2 = 16.0\text{ bar}$. The compression is carried out in two stages with an ideal intercooler between the stages that cools the air back to its inlet temperature. Determine the optimal intermediate pressure $P_i$ (in $\text{bar}$) that minimizes the total compression work.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (A) $W_3 > W_2 > W_1$ | Compressive Work Ordering ($V_0 \to V_0/2$) |
| Q2 | (A) $R T_0$ | Work Done in Isothermal Compression ($V_0/e$) |
| Q3 | (C) $1200\text{ K}$ | Adiabatic Compression Temperature Rise ($\gamma = 1.4$) |
| Q4 | (B) $\sqrt{P_1 P_2}$ | Optimal Intermediate Pressure for Intercooling |
| Q5 | (B) $2 P_1 V_1$ | Single-Stage Sudden Irreversible Work |
| Q6 | (A), (B), (C), (D) | Isothermal vs. Adiabatic Compression to $2P_0$ |
| Q7 | (A), (B), (C), (D) | Multi-Stage Intercooled Compressor Principles |
| Q8 | (A), (B), (C), (D) | First & Second Law of Irreversible Compression |
| Q9 | 9000 | Adiabatic Compressive Work ($4.5 \times 2000\text{ J}$) |
| Q10 | 4 | Geometric Mean Optimal Stage Pressure ($\sqrt{1 \times 16}\text{ bar}$) |
Complete Step-by-Step Solutions
Solution 1:
When compressing from $V_0$ to $V_0 / 2$, moving leftward on a $P$-$V$ indicator diagram:
– Isobaric path: $P = P_0$ (horizontal line, lowest curve).
– Isothermal path: $P \propto 1/V$, slope magnitude is $P/V$.
– Adiabatic path: $P \propto 1/V^\gamma$, slope magnitude is $\gamma P / V > P/V$.
Because the adiabatic curve lies above the isothermal curve, which lies above the isobaric line throughout $[V_0/2, V_0]$, the area under the curve (representing work done ON the gas) satisfies:
$$W_3 > W_2 > W_1$$
Correct Option: (A)
Solution 2:
Work done ON the gas during reversible isothermal compression from $V_0$ to $V_f = V_0 / e$:
$$W_{\text{on}} = n R T_0 \ln\left(\frac{V_0}{V_f}\right) = (1) R T_0 \ln\left(\frac{V_0}{V_0 / e}\right) = R T_0 \ln(e) = \mathbf{R T_0}$$
Correct Option: (A)
Solution 3:
For reversible adiabatic compression of a diatomic gas ($\gamma = 1.4 = 7/5$):
$$T_f V_f^{\gamma – 1} = T_0 V_0^{\gamma – 1} \implies T_f = T_0 \left(\frac{V_0}{V_f}\right)^{\gamma – 1}$$
Given $\frac{V_0}{V_f} = 32 = 2^5$ and $\gamma – 1 = 0.4 = 2/5$:
$$T_f = 300 \times (2^5)^{2/5} = 300 \times 2^2 = 300 \times 4 = \mathbf{1200\text{ K}}$$
Correct Option: (C)
Solution 4:
For two-stage compression with perfect intercooling:
$$W_{\text{total}} = \frac{n R T_1 \gamma}{\gamma – 1} \left[\left(\frac{P_i}{P_1}\right)^{\frac{\gamma – 1}{\gamma}} + \left(\frac{P_2}{P_i}\right)^{\frac{\gamma – 1}{\gamma}} – 2\right]$$
Minimizing $W_{\text{total}}$ with respect to $P_i$ by setting $\frac{dW_{\text{total}}}{dP_i} = 0$:
$$\frac{d}{dP_i}\left[P_i^k P_1^{-k} + P_2^k P_i^{-k}\right] = 0 \implies k P_i^{k-1} P_1^{-k} – k P_2^k P_i^{-k-1} = 0$$
$$P_i^{2k} = (P_1 P_2)^k \implies P_i^2 = P_1 P_2 \implies \mathbf{P_i = \sqrt{P_1 P_2}}$$
Correct Option: (B)
Solution 5:
For sudden single-stage compression against constant external pressure $P_{\text{ext}} = 3 P_1$:
Final volume by Boyle’s Law ($T = \text{const}$):
$$V_2 = \frac{P_1 V_1}{P_2} = \frac{P_1 V_1}{3 P_1} = \frac{V_1}{3}$$
Work done ON the gas:
$$W_{\text{on}} = P_{\text{ext}} (V_1 – V_2) = 3 P_1 \left(V_1 – \frac{V_1}{3}\right) = 3 P_1 \left(\frac{2 V_1}{3}\right) = \mathbf{2 P_1 V_1}$$
Correct Option: (B)
Solution 6:
(A) Correct: $V_{f, \text{iso}} = V_0 / 2 = 0.5 V_0$, while $V_{f, \text{adia}} = V_0 / 2^{1/\gamma} > 0.5 V_0$ (for monoatomic, $2^{3/5} \approx 1.516 \implies V_{f,\text{adia}} \approx 0.66 V_0$). Thus $V_{f, \text{A}} < V_{f, \text{B}}$.
(B) Correct: Compressing to $P_f = 2P_0$ requires more work along the adiabatic path because resisting pressure rises faster: $W_{\text{on, adia}} = \frac{P_0 V_0}{\gamma – 1}(2^{\frac{\gamma – 1}{\gamma}} – 1) > P_0 V_0 \ln 2$.
(C) Correct: $T_{f, \text{B}} = T_0 (2)^{\frac{\gamma – 1}{\gamma}} > T_0 = T_{f, \text{A}}$.
(D) Correct: On a $P$-$V$ diagram, the steeper adiabatic curve starts at $(P_0, V_0)$ and climbs above the isothermal curve throughout the compression.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: Cooling the gas between stages reduces its specific volume, shrinking the area under the $P$-$V$ curve in subsequent stages.
(B) Correct: Optimization requires $r_p = \sqrt{P_2 / P_1}$ for both stages.
(C) Correct: Equal pressure ratio and identical inlet temperature mean both stages perform identical work.
(D) Correct: In the limit of infinite stages with continuous intercooling, the path follows an isotherm: $W_{\text{min}} = n R T \ln(P_2 / P_1)$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: Internal energy of an ideal gas depends solely on temperature: $\Delta T = 0 \implies \Delta U = 0$ in both methods.
(B) Correct: By First Law, $Q = -W_{\text{on}}$. Since $W_{\text{on, II}} > W_{\text{on, I}}$, more heat is dumped to the reservoir in Method II: $|Q_{\text{II}}| > |Q_{\text{I}}|$.
(C) Correct: Entropy $S$ is a state function; since end states are identical, $\Delta S_{\text{gas}} = n R \ln(V_2 / V_1) < 0$ in both.
(D) Correct: Reversible path has $\Delta S_{\text{univ}} = 0$, while irreversible path generates entropy: $\Delta S_{\text{univ, II}} = \Delta S_{\text{gas}} + \frac{|Q_{\text{II}}|}{T_0} > 0$.
Correct Options: (A), (B), (C), (D)
Solution 9:
For reversible adiabatic compression of a monoatomic gas ($\gamma = 5/3, C_v = 1.5R$):
$$T_f = T_0 \left(\frac{V_0}{V_0 / 8}\right)^{\gamma – 1} = T_0 (8)^{2/3} = 4 T_0$$
Work done ON the gas equals the gain in internal energy:
$$W_{\text{on}} = \Delta U = n C_v (T_f – T_0) = \frac{3}{2} n R (4 T_0 – T_0) = \frac{3}{2} (3) n R T_0 = \frac{9}{2} n R T_0 = 4.5 \, n R T_0$$
Given $n R T_0 = 2000\text{ J}$:
$$W_{\text{on}} = 4.5 \times 2000\text{ J} = \mathbf{9000\text{ J}}$$
Answer: 9000
Solution 10:
For a two-stage compressor with perfect intercooling operating between $P_1 = 1.0\text{ bar}$ and $P_2 = 16.0\text{ bar}$, minimum total work requires the intermediate pressure to be the geometric mean:
$$P_i = \sqrt{P_1 P_2} = \sqrt{(1.0\text{ bar})(16.0\text{ bar})} = \sqrt{16} = \mathbf{4\text{ bar}}$$
Answer: 4